Day 12, Mat128: Calculus A

• Here's a key
• This was (essentially) the preview exercise from the section, so I was hoping that you'd all do well on it.

  (I did create the graphs myself, so I can show you the precise derivatives; I also added the "f." question.)

• There's still a disconnect between shapes of the derivative functions, and the shape of the function itself. In particular, you need to realize that this problem can be solved entirely with your ruler.

• In this case, since we're talking about the motion of a vehicle, each function has a precise meaning:
  0. \(s(t)\) -- position of the vehicle
  1. \(s'(t)\) -- velocity of the vehicle -- "\(v(t)\)"
     • Shows:
       1. Increase/decrease in the position function \(s(t)\)
       2. rate of change of the position function \(s(t)\)
       3. Speed is the size of the rate of change; direction is the sign of the rate of change

     • Use ruler slopes on \(s(t)\) to get \(s'(t)\) values

  2. \(s''(t)\) -- acceleration of the vehicle -- "\(v'(t)\)"
     • Shows:
       1. Concavity in the position function \(s(t)\)
       2. rate of change of the velocity function \(s'(t)\)
       3. acceleration of the position function \(s(t)\)

     • Use ruler slopes on \(s'(t)\) to get \(s''(t)\) values

• Important: Notice that the motion of the car is periodic: it does the same thing over and over, repeating every four minutes; hence its velocity graphs and acceleration graphs should also be periodic -- the same on each four minute interval.

• Let's review our example: suppose you need to compute $\sqrt{15.96}$. Now the easy way is to use your calculator, but the cowboy way is to use the linearization!
  • We know the square root of 16, which is close: our answer should be near 4. 16 becomes our \(a\) in \[ L(x)=f'(a)(x-a)+f(a)=f'(16)(x-16)+f(16) \]

  • Compute the derivative of $f(x)=\sqrt{x}$ at $x=16$, and we can write down the tangent line equation: \[ L(x) = \frac{1}{8}(x-16) + 4 \]

  • We use the tangent line for our approximation, and, when we're done, \[ L(15.96) = \frac{1}{8}(15.96-16) + 4 = \frac{1}{8}(-0.04) + 4 = -0.005 + 4 = 3.995 \approx 3.994996871087636 \] we can see why our answer an over-estimate:

    The answer takes us back to the second derivative:

    

    Here's a graph that Chris sent me, showing inflection at work!

    And here's one I showed him, in my work on climate change.

• Are there any questions about the worksheet (some activities and homework problems from section 1.8)?

• Theorem: Let $F(x)=c$ where $c$ is a constant. Then $F'(x)=0$.

  This one is obvious, "by slopes".

• Theorem: Let $F(x)=x$. Then $F'(x)=1$.

  This one is also obvious, "by slopes".

• Theorem (sum rule -- ``The derivative of a sum is the sum of the derivatives.'' It rolls off the tongue....):
  Let $F(x)=f(x)+g(x)$. \[ \frac{d}{dx}[F(x)] = \frac{d}{dx}[f(x)+g(x)] = \frac{d}{dx}f(x) + \frac{d}{dx}g(x) = f'(x) + g'(x) \]

• Theorem (product rule -- not simply the product of the derivatives! No tongue-rolling, please....):
  Let $F(x)=f(x)\cdot g(x)$. \[ \frac{d}{dx}[F(x)]=\frac{d}{dx}[f(x)\cdot g(x)]=f(x)\frac{d}{dx}g(x) + g(x)\frac{d}{dx}f(x) =f(x)g'(x) + g(x)f'(x) \]
• Corollary (constant multiple rule): multiplying a function by a constant scales it in the $y$ direction. This makes it steeper by a factor of $c$.
  Let $F(x)=cf(x)$. \[ \frac{d}{dx}[F(x)]=\frac{d}{dx}[cf(x)]=c\frac{d}{dx}f(x)=cf'(x) \] This is a consequence of the product rule and the constant rule (which is why we call it a corollary).

• Theorem (Power rule):
  Let $F(x)=x^n$, where $n$ is an integer, \(n \ge 1\). \[ \frac{d}{dx}[F(x)]=\frac{d}{dx}(x^n)=nx^{n-1} \]

  This proof will also make use of one of our new tools (the product rule). Once you build and prove a tool, it becomes a power tool:).

  The proof is actually "by dominoes" (or rather, the principle of mathematical induction....).

  The power works for any real exponent (except 0) -- it's just that we've only proven it for positive integers. You may use it for other powers (so write $\sqrt{x}=x^\frac{1}{2}$, for example, to make use of this nice new rule!).

• Theorem (exponential rule):
  Let $F(x)=a^x$, with \(a>0\) (but \(a \ne 1\)). \[ \frac{d}{dx}[F(x)] = \frac{d}{dx}[a^x] = \ln(a) a^x = \ln(a) F(x) \]

  This says that an exponential function has a slope function which is just a constant times the function itself (and the constant is the log, base \(e\), of \(a\)).

  In particular, let $F(x)=e^x$, where $e \approx 2.71828$ (it's irrational, like $\pi$). Then \[ \frac{d}{dx}[F(x)] = F(x) \hspace{1in} (= F'(x) = F''(x) = F'''(x) = ....) \]

  This is certainly one of the most amazing rules. It says that $e^x$ is an exponential function which is its own derivative: whose values are its slopes, as well (and all its higher derivatives as well -- its concavity, its jerk, ...).

  Here's an animation which motivates the discussion...

But, at the moment, we've got all the power we need to differentiate (easily) any polynomial, and any exponential function -- and that's a huge group of functions!